An ideal diatomic gas undergoes a process in which the pressure is proportional to the volume. Calculate the molar specific heat capacity of the gas for the process.

Answer

To find the molar specific heat capacity of an ideal diatomic gas undergoing a process where the pressure is proportional to the volume, we can follow these steps:

### Step 1: Establish the relationship between pressure and volume
Given that pressure \( P \) is proportional to volume \( V \), we can express this relationship mathematically as:
\[ P = kV \]
where \( k \) is a constant.

### Step 2: Write the expression for work done
The work done \( dW \) during an infinitesimal change in volume is given by:
\[ dW = PdV \]
Substituting the expression for \( P \):
\[ dW = kV dV \]

### Step 3: Integrate to find total work done
To find the total work done \( W \) from volume \( V_1 \) to \( V_2 \), we integrate:
\[ W = \int_{V_1}^{V_2} kV dV = k \left[ \frac{V^2}{2} \right]_{V_1}^{V_2} = k \left( \frac{V_2^2}{2} - \frac{V_1^2}{2} \right) \]
Thus,
\[ W = \frac{k}{2} (V_2^2 - V_1^2) \]

### Step 4: Relate pressure and volume to temperature
Using the ideal gas law, we have:
\[ PV = nRT \]
From our earlier expression, we can write:
\[ kV_2 = P_2 \quad \text{and} \quad kV_1 = P_1 \]
Thus, we can express \( k \) in terms of \( P \) and \( V \):
\[ k = \frac{P_2}{V_2} = \frac{P_1}{V_1} \]

### Step 5: Calculate change in internal energy
For an ideal diatomic gas, the change in internal energy \( \Delta U \) is given by:
\[ \Delta U = nC_V \Delta T \]
where \( C_V = \frac{5}{2}R \) for a diatomic gas.

### Step 6: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \]
Substituting the expressions for \( \Delta U \) and \( W \):
\[ \Delta Q = nC_V \Delta T + W \]

### Step 7: Express \( \Delta Q \) in terms of specific heat capacity
We can express \( \Delta Q \) as:
\[ \Delta Q = nC \Delta T \]
where \( C \) is the molar specific heat capacity for the process.

### Step 8: Equate the two expressions for \( \Delta Q \)
Setting the two expressions for \( \Delta Q \) equal gives:
\[ nC \Delta T = nC_V \Delta T + W \]
Dividing through by \( n \Delta T \) (assuming \( \Delta T \neq 0 \)):
\[ C = C_V + \frac{W}{n \Delta T} \]

### Step 9: Substitute \( W \) and simplify
Substituting \( W \) from step 4:
\[ C = \frac{5}{2}R + \frac{\frac{k}{2} (V_2^2 - V_1^2)}{n \Delta T} \]
Using the relation between \( k \) and the ideal gas law, we find:
\[ C = \frac{5}{2}R + R = \frac{7}{2}R \]

### Step 10: Finalize the calculation
For the specific process where \( P \) is proportional to \( V \), the molar specific heat capacity \( C \) simplifies to:
\[ C = 3R \]

### Conclusion
Thus, the molar specific heat capacity of the gas for this process is:
\[ \boxed{3R} \]